Problem #1
A human being can be electrocuted if a current as small as 50 mA passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding, one in each hand. If his resistance is 2000 Ω, what might the fatal voltage be?Answer;
Known:
the fatal current is ifatal = 50 mA = 50 x 10─3 A
The resistance of the human body is Rbody = 2000 Ω
If in a closed circuit, the current is given by I and the resistance is R, then the voltage is given by,
Vfatal = ifatalRbody = (50 x 10─3 A)(2000 Ω) = 100 V
Problem#2
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.0 A. The resistance per unit length is to be 0.150 Ω/km. The densities of copper and aluminum are 8960 and 2600 kg/m3, respectively. Compute (a) the magnitude J of the current density and (b) the mass per unit length λ for a copper cable and (c) J and (d) λ for an aluminum cable.
Answer;
Known:
transmission line that must carry a current of i = 60.0 A
The resistance per unit length is R/L = 0.150 Ω/km = 0.150 x 10─3 Ω/m
The densities of copper is dCu = 8960 kg/m3
The densities of aluminum is dAl = 2600 kg/m3
(a) the magnitude J of the current density a copper cable given by
J = i/A
the electrical resistance that there is an conductor is given by
R = ρL/A or A = ρL/R
then, J = i/A = iR/ρL
J = (60.0 A)(0.150 x 10─3 Ω/m)/(1.69 x 10─8 Ωm) = 5.33 x 105 A/m2
(b) the mass per unit length λ for a copper cable is given by
λ = m/L
while, m = dV,
λ = dV/L = dA
λ = ρd(L/R)
λ = (1.69 x 10─8 Ωm)(8960 kg/m3)(1/0.150) x 103 m/Ω = 1.01 kg/m
(c) the magnitude J of the current density a aluminum cable given by
J = iR/ρL = (60.0 A)(0.150 x 10─3 Ω/m)/(2.75 x 10─8 Ωm) = 3.27 x 105 A/m2
(d) the mass per unit length λ for a aluminum cable is given by
λ = ρd(L/R)
λ = (2.75 x 10─8 Ωm)(2600 kg/m3)(1/0.150) x 103 m/Ω = 0.477 kg/m
Problem #3
A wire of Nichrome (a nickel–chromium–iron alloy commonly used in heating elements) is 1.0 m long and 1.0 mm2 in cross-sectional area. It carries a current of 4.0 A when a 2.0 V potential difference is applied between its ends. Calculate the conductivity σ of Nichrome.
Answer;
Known:
Current, i = 4.0 A
Potential difference, V = 2.0 V
Wire of Nichrome of length is L = 1.0 m
Cross-sectional area is A = 1.0 mm2 = 1.0 x 10─6 m2
conductivity σ of Nichrome given by
σ = 1/ρ
with R = ρL/A and R = V/i, then,
ρ = VA/iL
so that
σ = iL/VA
= (4.0 A)(1.0 m)/(2.0 V x 1.0 x 10─6 m2)
σ = 2.0 x 106 Ω─1m─1
Problem#4
A wire 4.00 m long and 6.00 mm in diameter has a resistance of 15.0 mΩ. A potential difference of 23.0 V is applied between the ends. (a) What is the current in the wire? (b) What is the magnitude of the current density? (c) Calculate the resistivity of the wire material. (d) Identify the material.
Answer;
Known:
A wire with length of L = 4.00 m
A wire with length of d = 6.00 mm = 6.00 x 10─3 m
Resistance of R = 15.0 mΩ = 15.0 x 10─3 Ω
A potential difference of V = 23.0 V
(a) the current in the wire?
i = V/R = 23.0 V/(15.0 x 10─3 Ω) = 1533.33 A
(b) the magnitude of the current density given by
J = i/A
Where A is the cross─sectional area of the wire, and its given by
A = πr2 = πd2/4
Subtitute to get:
J = 4i/πd2 = 4 x (1533.33 A)/[π x (6.00 x 10─3 m)2] = 5.42 x 107 A/m2
(c) the resistance for the given wire is:
R = ρL/A or rewrite ρ = RA/L
Subtitute with the given values to get:
ρ = πRd2/4L
= π(15.0 x 10─3 Ω)(6.00 x 10─3 m)2/(4 x 4.00 m)
ρ = 1.06 x 10─7 Ωm
(d) from part (c) the resistivity belong to the platinum, so material is platinum
Problem#5
A common flashlight bulb is rated at 0.30 A and 2.9 V (the values of the current and voltage under operating conditions). If the resistance of the tungsten bulb filament at room temperature (200C) is 1.1 Ω, what is the temperature of the filament when the bulb is on?
Answer;
Known:
Current, i = 0.30 A
Potential difference, V = 2.9 V
The resistance of the tungsten bulb filament at room temperature (T0 = 200C) is R0 = 1.1 Ω
the temperature of the filament when the bulb is on,
R = V/i = 2.9 V/0.30 A = 9.67 Ω
We have:
R = R0[1 + α(T ─ T0)]
9.67 Ω = 1.1 Ω[1 + 4.5 x 10─3 Ω/0C(T ─ 200)]
T = 18000C
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